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Question: The value of \(\sin {1^0} + \sin {2^0} + \sin {3^0} + ...... + \sin {359^0}\) is a. 0 b. 1 c. ...

The value of sin⁡10+sin⁡20+sin⁡30+......+sin⁡3590\sin {1^0} + \sin {2^0} + \sin {3^0} + ...... + \sin {359^0} is
a. 0
b. 1
c. -1
d. 180

Explanation

Solution

Hint: Here, we will solve the given equation by checking whether the number of terms is even or not and applying the relevant trigonometric formulae.

Complete step-by-step answer:
As you know that
sin⁡(3600−θ)=−sin⁡θ→(1)\sin ({360^0} - \theta ) = - \sin \theta \to (1)
⇒sin⁡10+sin⁡20+sin⁡30+.......sin⁡3590 ⇒sin⁡10+sin⁡20+sin⁡30+.......sin⁡3570+sin⁡3580+sin⁡3590 ⇒sin⁡10+sin⁡20+sin⁡30+.......sin⁡(3600−30)+sin⁡(3600−20)+sin⁡(3600−10)  \Rightarrow \sin {1^0} + \sin {2^0} + \sin {3^0} + .......\sin {359^0} \\\ \Rightarrow \sin {1^0} + \sin {2^0} + \sin {3^0} + .......\sin {357^0} + \sin {358^0} + \sin {359^0} \\\ \Rightarrow \sin {1^0} + \sin {2^0} + \sin {3^0} + .......\sin ({360^0} - {3^0}) + \sin ({360^0} - {2^0}) + \sin ({360^0} - {1^0}) \\\
From equation 1
⇒sin⁡10+sin⁡20+sin⁡30+.......−sin⁡30−sin⁡20−sin⁡10→(2)\Rightarrow \sin {1^0} + \sin {2^0} + \sin {3^0} + ....... - \sin {3^0} - \sin {2^0} - \sin {1^0} \to (2)
Now you know number of terms in series (1,2,3............359)(1,2,3............359) is even
From equation 2
All terms will cancel out
⇒0+0+0+0+..............+0=0\Rightarrow 0 + 0 + 0 + 0 + .............. + 0 = 0
Therefore option ‘a’ is correct.

Note: In this type of questions always check whether the number of terms is even or not and always remember the trigonometry properties. It is also crucial to know the properties of sine function in different quadrants.